Openness of $S=\{(x, \sin(1/x) : x>0\}$ Let $S=\{(x, \sin(1/x) : x>0\}$.
My understanding is that $S$ is not open in $\mathbb{R}^2$, but $S$ is open in the subspace topology.
Is it correct?
Thank you.
 A: Yes, you are right.  Do you understand why you are right?
Here is an intuitive idea of why you are right about $S$ not being open:
Think about the graph of $f(x) = \sin(1/x)$ in the $xy$-plane for $x > 0$.  Now, pick some point on the curve, maybe at $x = 100$ (or anywhere).  Draw a ball around that point of any radius $\epsilon$.  Are only elements of that graph in the ball?  No, of course not.  You will always have pieces of "white space" (i.e., the complement of $S$) in the open ball around the point in $S$, and this is true for any ball you draw, of any radius.  So $S$ can't be open, since if it were open, then for each point in it, we can find a radius such that the ball around that point contains only points in $S$.
Now, as far as $S$ being open in the subspace topology, you have to specify which subspace you are talking about.  If you mean the subspace topology of $S$, then of course $S$ is open in its own subspace topology.  That's a trivial statement, like saying $\Bbb R$ is open in its own topology.  Intuitively, if you pick any point on $S$, then the "ball" in the subspace topology looks like an open interval, but it is an open interval on the curve itself.  Of course, any point on the curve has an open ball around it that contains only points in the curve (if we are working in the subspace topology).
